1 Answer

Craastad · Answer 1 · 2022-06-18T11:56:20+0000

Answer:

See Explanation

Explanation:

The question is incomplete, as a table or chart that shows the required data is not given.

To answer this, I will make use of the following frequency table

$\begin{array}{cccccc}{Miles} & {1} & {1(1)/(2)} & {2} & {2(1)/(2)} & {3} \ \\ {Students} & {10} & {0} & {5} & {10} & {5} \ \end{array}$

From the above table.

$2(1)/(2)\ miles \to 10\ students$

$3\ miles \to 5\ students$

So:

$2(1)/(2)\ or\ 3\ miles = 10+5$

$2(1)/(2)\ or\ 3\ miles = 15\ students\\$

From the above table.

$1\ miles \to 10\ students$

$1(1)/(2)\ miles \to 0\ students$

So:

$1(1)/(2)\ or\ 1\ miles = 0+10$

$1(1)/(2)\ or\ 1\ miles = 10\ students$

The difference (d) is then calculated by subtracting the number of students in both categories

$d = 15\ students - 10\ students$

$d = 5\ students$

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